To solve the problem, we begin by examining the vector $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$.

Let's calculate:

$\vec{c} \cdot \hat{a} = 3 + 6(\hat{a} \cdot \hat{b})$.

$\vec{c} \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6$.

We need to find the value of $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$.

Substituting the expressions, we get:

$ 9(\vec{c} \cdot \hat{a}) = 9(3 + 6\hat{a} \cdot \hat{b}) = 27 + 54(\hat{a} \cdot \hat{b}) $

$ 3(\vec{c} \cdot \hat{b}) = 3(3\hat{a} \cdot \hat{b} + 6) = 9\hat{a} \cdot \hat{b} + 18 $

Therefore,

$ 9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = (27 + 54(\hat{a} \cdot \hat{b})) - (9\hat{a} \cdot \hat{b} + 18) $

Simplifying,

$ = 27 - 18 + 54(\hat{a} \cdot \hat{b}) - 9(\hat{a} \cdot \hat{b}) $

$ = 9 + 45(\hat{a} \cdot \hat{b}) $

Given $\sin{\theta} = \frac{\sqrt{65}}{9}$, and knowing that for unit vectors $\cos{\theta} = \hat{a} \cdot \hat{b}$, we use the identity $(\cos{\theta})^2 = 1 - (\sin{\theta})^2$:

$ (\hat{a} \cdot \hat{b})^2 = 1 - \left(\frac{\sqrt{65}}{9}\right)^2 $

$ = 1 - \frac{65}{81} $

$ = \frac{16}{81} $

Thus, $\hat{a} \cdot \hat{b} = \frac{4}{9}$.

Substitute back into the equation:

$ 9 + 45 \times \frac{4}{9} = 9 + 20 = 29 $

Therefore, the value is 29.